A long straight wire AB carries current 10 A. A proton (P) travels with a speed \( 2\times10^{6}\,\mathrm{m\,s^{-1}} \), parallel to the wire, at a distance 2 m from it in a direction opposite to the current as shown in figure. The force experienced by the proton is: \( (\mu_0 = 4\pi\times10^{-7}\,\mathrm{N\,A^{-2}}) \)
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \( 3.2\times10^{-19}\,\mathrm{N} \)
The wire produces a magnetic field at the proton, and the moving proton experiences the magnetic force \( F = qvB\sin\theta \).
\( \displaystyle B = \frac{\mu_0I}{2\pi r} = \frac{4\pi\times10^{-7}\times10}{2\pi\times2} = 1\times10^{-6}\,\mathrm{T} \)
The field lines are circles around the wire, so at the proton \( \vec{B} \) is perpendicular to the wire, while \( \vec{v} \) is parallel to it. Hence \( \theta = 90^\circ \).
\( F = qvB = 1.6\times10^{-19}\times2\times10^{6}\times1\times10^{-6} = 3.2\times10^{-19}\,\mathrm{N} \)
The proton moving opposite to the current is like a current antiparallel to the wire's current, so the force on it is directed away from the wire.